# Is equality of cardinality an equivalence relation on sets?

I am trying to show that equality of cardinality is an equivalence relation.

I am a bit confused as to how I should approach this.

Currently, my understanding of equivalence relation is that a relation needs to be

reflexive, symmetric, and transitive in order to be an equivalence relation.

For instance, if I have

$$A = \{1,2,3,5\}$$

$$R = \{(1,1),(2,2),(3,3),(5,5),(1,5),(5,1)\}$$, $$R \subseteq A \times A$$

$$R$$ would be an equivalence relation.

If relation $$R$$ is an equality of cardinality, I am not sure as to how I can

show that it is reflexive, symmetric, and transitive.

Could somebody help me?

• An equivalence relation is a relation on a single set $A$, e.g. a subset of the Cartesian product $A \times A$. Your example fails to be an equivalence relation because it is not a relation on $A \times A$ or $B\times B$, but on $A\times B$. – Xander Henderson Oct 20 '19 at 23:23
• Welcome to Maths SX! How do you define equal cardinalities? – Bernard Oct 20 '19 at 23:36
• Your example is not an equivalence relation because it is not reflexive. $2 \not R2$ for example. It is symmetric and transitive. – Ross Millikan Oct 20 '19 at 23:50

## 1 Answer

Hint. By definition, two sets $$A$$ and $$B$$ have the same cardinality, hence $$A\sim B$$, if there exists a bijection $$A\to B$$. On the other hand if we have a bijection $$A\to B$$ and $$B\to C$$, we need to show there is a bijection $$A\to C$$, which would mean that $$A\sim B$$ and $$B\sim C$$ implies $$A\sim C$$ (transitivity); can you try to do so using the fact that composition of bijections is a bijection? To show it is symmetric, given a bijection $$A\to B$$ we need to find a bijection $$B\to A$$, which should be easy to show, noting the fact that bijections have inverses which are also bijections. Finally to see $$A\sim A$$ (reflexivity) we just need to find a bijection from $$A$$ to itself, which should be easy to see.